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Math Help - Prove there is a number a

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    Prove there is a number a

    Suppose f: D --> R is a function whose domain contains [a,b] for some a,b \in R amd f is integrable on [a,b]. suppose a < c < b.

    1. Suppose A is a set of real number that is bounded above. Prove that for any \epsilon > 0 there is a number a \inA Such that supA-a < \epsilon.
    (Hint: Prove by contradiction. If there is an \epsilon >0 such that supA-a >= \epsilon for all a \in A then what does this say about the number supA - \epsilon ?)
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    Member Abu-Khalil's Avatar
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    Suppose there exists \epsilon>0 such as \sup A-a\geq \epsilon,\forall a\in A\Rightarrow \sup A-\epsilon \geq a,\forall a\in A\Rightarrow \exists b<\sup A:b\geq a,\forall a\in A. Hence, \sup A isn't the lower upper bound.
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    Quote Originally Posted by Abu-Khalil View Post
    Suppose there exists \epsilon>0 such as \sup A-a\geq \epsilon,\forall a\in A\Rightarrow \sup A-\epsilon \geq a,\forall a\in A\Rightarrow \exists b<\sup A:b\geq a,\forall a\in A. Hence, \sup A isn't the lower upper bound.
    Can you explain what we are proving here, supA - a < \epsilo implies it is a least upper bound or not a least upper bound?
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    MHF Contributor Drexel28's Avatar
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    Quote Originally Posted by 450081592 View Post
    Suppose f: D --> R is a function whose domain contains [a,b] for some a,b \in R amd f is integrable on [a,b]. suppose a < c < b.

    1. Suppose A is a set of real number that is bounded above. Prove that for any \epsilon > 0 there is a number a \inA Such that supA-a < \epsilon.
    (Hint: Prove by contradiction. If there is an \epsilon >0 such that supA-a >= \epsilon for all a \in A then what does this say about the number supA - \epsilon ?)
    This is really quite simple my good man. Suppose for a second that there existed some \varepsilon>0 such that no element of A was greater than \sup\text{ }A-\varepsilon. Then, it must be an upper bound of A but since \text{sup }A-\varepsilon<\text{sup }A this contradicts \sup\text{ }A being the least upper bound.
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    Quote Originally Posted by Drexel28 View Post
    This is really quite simple my good man. Suppose for a second that there existed some \varepsilon>0 such that no element of A was greater than \sup\text{ }A-\varepsilon. Then, it must be an upper bound of A but since \text{sup }A-\varepsilon<\text{sup }A this contradicts \sup\text{ }A being the least upper bound.
    Ya I am clear now, thanks a lot
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